2 years ago

50 points :) 100-70=?

Mathematics

2 answers:

Marina86 [1]2 years ago

3 0

Answer:

Step-by-step explanation:

Did you just do this for fun?

Either way have a good rest of your day!

abruzzese [7]2 years ago

3 0

Answer:

THANK YOU!!!

I really appreciate this have a great day

You might be interested in

Determine the relationship between the two given lines y=3 and y=2

nekit [7.7K]

They’re parallel to each other

3 0

2 years ago

2v+7=3 Help me out I’m dating vinnie hacker

Mariana [72]

2v=3-7
2v=-4 divide both sides
v=-2

4 0

2 years ago

Read 2 more answers

Help me PLZ with the exact answer

olchik [2.2K]

Y = 4/-3 - 1

The slope can either be -4/3 or 4/-3 and the y-intercept is (0,1)

4 0

3 years ago

PLSSS HELP IF YOU TURLY KNOW THISS

Phoenix [80]

Answer:

2x + 3y +1

Step-by-step explanation:

-2x + 4x -6y + 9y +3 -2 (arranging)

2x + 3y + 1

8 0

2 years ago

Read 2 more answers

Consider the following equation. f(x, y) = y3/x, P(1, 2), u = 1 3 2i + 5 j (a) Find the gradient of f. ∇f(x, y) = Correct: Your

BaLLatris [955]

$f(x,y)=\dfrac{y^3}x$

a. The gradient is

$\nabla f(x,y)=\dfrac{\partial f}{\partial x}\,\vec\imath+\dfrac{\partial f}{\partial y}\,\vec\jmath$

$\boxed{\nabla f(x,y)=-\dfrac{y^3}{x^2}\,\vec\imath+\dfrac{3y^2}x\,\vec\jmath}$

b. The gradient at point P(1, 2) is

$\boxed{\nabla f(1,2)=-8\,\vec\imath+12\,\vec\jmath}$

c. The derivative of $f$ at P in the direction of $\vec u$ is

$D_{\vec u}f(1,2)=\nabla f(1,2)\cdot\dfrac{\vec u}{\|\vec u\|}$

It looks like

$\vec u=\dfrac{13}2\,\vec\imath+5\,\vec\jmath$

so that

$\|\vec u\|=\sqrt{\left(\dfrac{13}2\right)^2+5^2}=\dfrac{\sqrt{269}}2$

Then

$D_{\vec u}f(1,2)=\dfrac{\left(-8\,\vec\imath+12\,\vec\jmath\right)\cdot\left(\frac{13}2\,\vec\imath+5\,\vec\jmath\right)}{\frac{\sqrt{269}}2}$

$\boxed{D_{\vec u}f(1,2)=\dfrac{16}{\sqrt{269}}}$

7 0

3 years ago